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If P, Q Are Real And P ≠ Q, Then Show that the Roots of the Equation (P − Q) X2 + 5(P + Q) X− 2(P − Q) = 0 Are Real and Unequal. - CBSE Class 10 - Mathematics

Question

If p, q are real and p ≠ q, then show that the roots of the equation (p − q) x2 + 5(p + q) x− 2(p − q) = 0 are real and unequal.

Solution

The quadric equation is (p − q) x2 + 5(p + q) x− 2(p − q) = 0

Here,

a = (p - q), b = 5(p + q) and c = -2(p - q)

As we know that D = b2 - 4ac

Putting the value of a = (p - q), b = 5(p + q) and c = -2(p - q)

D = {5(p + q)}2 - 4 x (p - q) x (-2(p - q))

= 25(p2 + 2pq + q2) + 8(p2 - 2pq + q2)

= 25p2 + 50pq + 25q2 + 8p2 - 16pq + 8q2

= 33p2 + 34pq + 33q2

Since, P and q are real and p ≠ q, therefore, the value of D ≥ 0.

Thus, the roots of the given equation are real and unequal.

Hence, proved

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Solution If P, Q Are Real And P ≠ Q, Then Show that the Roots of the Equation (P − Q) X2 + 5(P + Q) X− 2(P − Q) = 0 Are Real and Unequal. Concept: Nature of Roots.
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